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Solving First Order Linear Differential Equations
Example 1. A 20-quart juice dispenser in a cafeteria is filled with a juice
mixture that is 10% cranberry and 90% orange juice. A pineapple-orange blend
(40% pineapple and 60% orange) is entering the dispenser at a rate of 4 quarts
an hour and the well-stirred mixture leaves at a rate of 5 quarts an hour. Model
the situation with a differential equation whose solution is the amount of orange
juice in the container at time t.
Let y = y(t) be the amount of orange juice in the container at time t.
Notice that unlike previous problems we’ve done, the rate in and rate out
are not the same. Each hour the number of gallons of juice mixture decreases
by 1 quart.
dy = rate in of orange juice - rate out of orange juice
dt
rate in = 4 qts mixture · (0.6) qts orange = 2.4 qts orange
hr qts mixture hr
rate out = 5 qts mixture · y qts orange = 5y qtsorange
hr (20−t)qts mixture 20−t hr
The differential equation modelling the situation is
dy =2.4− 5y
dt 20−t
where y(0) = 18.
This differential equation is not separable. But it is a first order linear dif-
ferential equation and by the end of this handout you should be able to solve
it.
Definition: A first order linear differential equation is a differential
equation that can be put in the form
dy +P(x)y =Q(x).
dx
Wewill refer to this as ‘standard form’ for such a differential equation.
The goal of this handout is to show you how to systematically solve such a
differential equation. The next example serves to motivate the method.
dy 1 2
Example 2. Solve dx + xy = x .
First verify for yourself that this differential equation is not separable. Having
established that, the prospect of solving might look grim. However, suppose we
were to multiply both sides of the equation by x.
dy 1 2
dx + xy = x
dy 3
xdx +y = x
1
If you look long and hard at the left hand side of the equation above you might
start to see the product rule in there and recognize that the left hand side of
this equation is the derivative of xy with respect to x. That observation proves
to be very handy because we can then integrate with respect to x and solve for
y as follows:
d 3
dx(xy) = x
Z d Z 3
dx(xy) dx = x dx
4
xy = x +C
4
3
y = x +C
4 x
Note that we cannot replace C by a constant C because x is not a constant!
x 1
What saved the day was the idea of multiplying both sides of the differential
equation by some function so that the left hand side of the differential equation
was just the derivative of a product. Let’s try to do this in general. (The
rewards are great – we’ll end up with a recipe for solving this type of differential
equation.)
General Case: We begin with
dy +P(x)y =Q(x).
dx
Our goal is to find a function v(x) such that when we multiply both sides by
v(x) the left hand side looks like a product. Let v(x) be a positive function.
Multiplying by v(x) gives
v(x)dy +v(x)P(x)y = v(x)Q(x). (1)
dx
Thinking about the product rule we see that we should aim for a function v(x)
such that the left hand side of (1) is d (vy). In other words, we want to find
v(x) such that dx
v(x)dy +v(x)P(x)y = d (vy). (2)
dx dx
From the product rule we know that d (vy) = v(x)dy + dv y so our goal is to
find v(x) such that dx dx dx
v(x)dy +v(x)P(x)y = v(x)dy + dvy.
dx dx dx
This amounts to finding v(x) such that
v(x)P(x)y = dvy, or equivalently v(x)P(x) = dv.
dx dx
2
The latter is a separable differential equation. We solve for v(x) as shown
below. Keep in mind that in any particular problem P(x) is known.
v(x)P(x) = dv
dx
P(x)dx = 1dv
v
Z P(x)dx = Z 1dv
Z v
P(x)dx = lnv
R P(x)dx
v = e
R P(x)dx
v = e is called the integrating factor. We now know that we can
indeed find v such that the left hand side of (1) is just d (vy) we can proceed
as follows. dx
v(x)dy +v(x)P(x)y = v(x)Q(x)
dx
d vy = v(x)Q(x)
dx Z
vy = v(x)Q(x)dx+C
y = 1 Z v(x)Q(x)dx+C
v
R P(x)dx
where v(x) = e .
Wehave a systematic way to solve first order linear differential equations.
• Step 1: Put the equation into standard form. Identify P(x) and Q(x).
R P(x)dx 1
• Step 2: Find the integrating factor v(x) = e and simplify.
• Step 3: Multiplying by v and integrating both sides gives
v(x)y = Z v(x)Q(x)dx+C.
If you really want a formula, here it is:
y = 1 Z v(x)Q(x)dx+C .
v
1When you integrate P(x) you can let the constant be zero as you’re just looking for one
integrating factor.
3
If the equation is actually separable, it is advisable to simply separate variables!
dy 2x
Example 3. Solve dx = e +3y. Find the general solution and then the
particular solution with y(0) = 3.
dy 2x
This is not separable, so we put it in standard form: dx − 3y = e .
2x
P(x) = −3 and Q(x) = e
R P(x)dx R −3dx −3x
v(x) = e =e =e . Multiplying by v(x) gives
−3xdy −3x −x
e dx −3e y = e
d −3x −x
dx(e y) = e
−3x Z −x
e y = e dx+C
−3x −x
e y = −e +C
2x 3x
y = −e +Ce
0 0 2x 3x
If y(0) = 3, then 3 = −e +Ce = −1+C, so C = 4 and y = −e +4e .
Note that if you really wanted to, after identifying P, Q, and v, you could skip
d −3x −x
directly to dx(e y) = e dx+C and solve.
2 ′
Example 4. Solve (x +1)y +3xy = 6x.
This is not separable, so we put it in standard form: y′ + 3x y = 6x .
x2+1 x2+1
P(x) = 3x and Q(x) = 6x .
2 2
x +1 R x +1
R P(x)dx 3x dx 2
2 3/2ln(x +1) 2 3/2
v(x) = e =e x +1 =e =(x +1) .
2 3/2 Z 2 3/2 6x
(x +1) y = (x +1) 2 dx+C
Z x +1
2 3/2 2 1/2
(x +1) y = 6x(x +1) dx+C
2 3/2 2 3/2
(x +1) y = 2(x +1) +C
2 −3/2 2 3/2
y = 2(x +1) [(x +1) +C]
y = 2+ C
2 3/2
(x +1)
Now you are ready to go back and solve the equation from Example 1.
Homework problems: Solve the following:
1. y′ − 4xy = x
′ y 3
2. y −3x +x −x=0
′ x
3. y +y = e where y(0) = 6.
4
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